Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define f first
#define s second
#define mp make_pair
#define inf 1<<30
using namespace std;

vector<int> v[100001];
int n,m,s,t,d1[100001],d2[100001];

void dijkstra1(){
  for(int i=0;i<n;i++)d1[i]=inf;
  priority_queue<pair<int,int> > pq;
  d1[s]=0;
  pq.push(mp(0,s));
  while(!pq.empty()){
    pair<int,int> u=pq.top();
    pq.pop();
    int a=(-1)*u.f,b=u.s;

    if(d1[b]<a)continue;

    for(int i=0;i<v[b].size();i++){
      if(a+1<d1[v[b][i]]){
	d1[v[b][i]]=a+1;
	pq.push(mp((-1)*(a+1),v[b][i]));
      }
    }
  }
}

void dijkstra2(){
  for(int i=0;i<n;i++)d2[i]=inf;
  priority_queue<pair<int,int> > pq;
  d2[t]=0;
  pq.push(mp(0,t));
  while(!pq.empty()){
    pair<int,int> u=pq.top();
    pq.pop();
    int a=(-1)*u.f,b=u.s;

    if(d2[b]<a)continue;

    for(int i=0;i<v[b].size();i++){
      if(a+1<d2[v[b][i]]){
	d2[v[b][i]]=a+1;
	pq.push(mp((-1)*(a+1),v[b][i]));
      }
    }
  }
}

int main()
{
  int a,b;

  cin>>n>>m>>s>>t;
  s--;t--;
  for(int i=0;i<m;i++){
    cin>>a>>b;
    a--;b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dijkstra1();
  dijkstra2();

  long long A[100001]={},B[100001]={},minc=d1[t],ans=0;

  for(int i=0;i<100001;i++){
    A[i]=0;
    B[i]=0;
  }
  for(int i=0;i<n;i++){
    if(d1[i]==inf)continue;
    A[d1[i]]++;
    B[d2[i]]++;
  }
  for(int i=0;i<d1[t]-1;i++){
    ans+=A[i]*B[minc-2-i];
  }
  cout<<ans<<endl;
  return 0;
}